Question

A challenging derivative Find \(d y / d x,\) where \(\sqrt{3 x^{7}+y^{2}}=\sin ^{2} y+100 x y\)

Short answer

#Question# Find the derivative \(\frac{dy}{dx}\) of the function \( \sqrt{3x^7 + y^2} = \sin^2y + 100xy\). #Answer# \(\frac{dy}{dx} = \frac{100y - \frac{21x^6}{2\sqrt{3x^7 + y^2}}}{2y \cdot \frac{1}{2\sqrt{3x^7 + y^2}} - 2\sin y \cos y - 100x}\)

Step-by-step solution

Differentiate both sides of the equation with respect to x

First, we have to differentiate both sides of the given equation. Remember to apply the chain rule when differentiating: \(\frac{d}{dx}(\sqrt{3x^7 + y^2}) = \frac{d}{dx}(\sin^2y + 100xy)\)

Apply the chain rule on the left side

The left side of the equation consists of a square root. Apply the chain rule by differentiating the square root and then multiplying by the derivative of the inside: \(\frac{1}{2\sqrt{3x^7 + y^2}} \cdot \frac{d}{dx}(3x^7 + y^2)\) Now differentiate the inside: \(\frac{1}{2\sqrt{3x^7 + y^2}} \cdot (21x^6 + 2y \cdot \frac{dy}{dx})\)

Apply the chain rule on the right side

The right side of the equation consists of a sine function squared and a product of two functions: \(\frac{d}{dx}(\sin^2y + 100xy) = 2\sin y \cdot \frac{d}{dx}(\sin y) + 100y + 100x \cdot \frac{dy}{dx}\) Now differentiate the inside function (sin y) using chain rule: \(2\sin y \cdot (\cos y \cdot \frac{dy}{dx}) + 100y + 100x \cdot \frac{dy}{dx}\)

Equate the left and right sides

Now that we have differentiated both sides, equate them to each other: \(\frac{1}{2\sqrt{3x^7 + y^2}} \cdot (21x^6 + 2y \cdot \frac{dy}{dx}) = 2\sin y \cdot (\cos y \cdot \frac{dy}{dx}) + 100y + 100x \cdot \frac{dy}{dx}\)

Solve for \(\frac{dy}{dx}\)

Now, isolate all the terms containing \(\frac{dy}{dx}\) on one side and move the other terms to the other side: \(\frac{dy}{dx}(2y \cdot \frac{1}{2\sqrt{3x^7 + y^2}} - 2\sin y \cos y - 100x) = 100y - \frac{21x^6}{2\sqrt{3x^7 + y^2}}\) Divide both sides by the terms in parenthesis: \(\frac{dy}{dx} = \frac{100y - \frac{21x^6}{2\sqrt{3x^7 + y^2}}}{2y \cdot \frac{1}{2\sqrt{3x^7 + y^2}} - 2\sin y \cos y - 100x}\) Now, you have the derivative of the function with respect to x as follows: \(\frac{dy}{dx} = \frac{100y - \frac{21x^6}{2\sqrt{3x^7 + y^2}}}{2y \cdot \frac{1}{2\sqrt{3x^7 + y^2}} - 2\sin y \cos y - 100x}\)

Key concepts

Chain Rule

When it comes to a complex function, the chain rule is the Swiss Army knife of differentiation. It allows us to take the derivative of a function that is composed of other functions, which is exactly what we encounter when dealing with nested expressions, such as a square root or a trigonometric function raised to a power.

The general concept is to 'unpack' the nested functions one by one. This process is akin to peeling an onion, layer by layer. As we take the derivative of the outer function, we must not forget to multiply by the derivative of the inner function. For instance, if we have a function \(u(v(x))\), the derivative \(\frac{du}{dx}\) is \(\frac{du}{dv}\times\frac{dv}{dx}\). In more intuitive terms, it's akin to asking: how does 'u' change as 'v' changes, and how does 'v' change as 'x' changes? Multiplying these rates together shows us how 'u' changes with respect to 'x', directly.

In the given problem, we applied the chain rule both to the square root function and the squared sine function, where the inner functions were \(3x^7 + y^2\) and \(\sin y\), respectively. The chain rule allowed us to dismantle these layers correctly and find how the overall expression changes with 'x'.

Derivative of Trigonometric Functions

Trigonometric functions are not only fundamental in geometry but also in calculus, where they often appear in various forms. Finding the derivative of trigonometric functions involves understanding their rates of change. Each trigonometric function has a specific derivative 'recipe'.

For instance, the sine function, \(\sin x\), has a derivative of \(\cos x\), indicating how the sine wave slopes upward and downward.

When Squares Come Into Play

The exercise deals specifically with \(\sin^2 y\), which means \(\sin y\) multiplied by itself. In such a case, we must apply both the chain rule and the derivatives of trigonometric functions. We differentiate the outer function, which is the square, resulting in \(2\sin y\), and then multiply by the derivative of the inner function, \(\sin y\), leading to \(\cos y\). The result is the product \(2\sin y \cdot \cos y\), showing the rate at which \(\sin^2 y\) changes with respect to 'y'. Then, considering the input 'y' is itself a function of 'x', which makes a further application of the chain rule necessary.

This detail in trigonometric differentiation is vital for solving problems concerning oscillatory motion, wave phenomena, and even complex electrical engineering problems.

Product Rule

The product rule is incredibly handy when tackling derivatives of expressions where two differentiable functions are multiplied together. This rule states that to find the derivative of a product of two functions, \(u\) and \(v\), you take the derivative of \(u\) and multiply it by \(v\), then add the product of \(u\) and the derivative of \(v\). Mathematically, it is written as \(\frac{d}{dx}(uv) = u'v + uv'\).

To break this down, it’s like saying that the rate of change of the whole is the sum of each part's contribution, taking turns at growth while the other remains constant.

Applying the Product Rule

In our example, the product of \(100xy\) required us to use the product rule. We took the derivative of \(x\), which is simply 1, and multiplied by \(y\), and then added the product of \(x\) and the derivative of \(y\) (which is \(\frac{dy}{dx}\)). This action perfectly illustrates how intertwined the functions are, and how their individual changes impact the rate at which their product changes.

Derivative of Exponential Functions

Exponential functions, expressed as \(a^x\) where \(a\) is a constant, grow or decay at a rate proportional to their current value. The beauty of the natural exponential function, \(e^x\), lies in its differentiation simplicity: its derivative is itself, \(\frac{d}{dx}e^x = e^x\).

For other bases, such as \(a^x\), the derivative introduces a factor of \(\ln(a)\), giving us \(\frac{d}{dx}a^x = a^x \ln(a)\). This constant, \(\ln(a)\), represents the rate at which the function \(a^x\) scales as 'x' changes.

Exponential Growth and Decay

While this concept didn’t directly come up in the given problem, understanding how to differentiate exponential functions is crucial. For example, in finance and biology, exponential growth or decay models processes such as compound interest or population dynamics. While these models might seem distant from our current problem's context, they share the key idea that to understand complex systems, we often break down the phenomena into their infinitesimal changes and analyze their behavior under the lens of calculus.